hd 2141 Can you find it?（二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 23753    Accepted Submission(s): 6012


Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth
line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

 

Sample Output

Case 1:
NO
YES
NO

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
long long a[510],b[510],c[510],ans[510*510];
int main()
{
int l , n ,m , i ,k , j , falg = 1 , s;
while(~scanf("%d%d%d",&l,&n,&m))
{
int g = 0;
for( i = 1 ; i <= l ; i++)
scanf("%lld",&a[i]);
for( j = 1 ; j <= n ; j++)
scanf("%lld",&b[j]);
for( k = 1 ; k <= m ; k++)
scanf("%lld",&c[k]);
for( i = 1 ; i <= l ; i++)
for(j = 1 ; j <= n ; j++){
ans[++g] = a[i] + b[j];
}
sort(ans+1,ans+g+1);
scanf("%d",&s);
printf("Case %d:\n",falg++);
while(s--)
{
long long num;
bool node = false;
scanf("%lld",&num);
for( i = 1 ; i <= m ; i++)
{
int l = 1 , r = g;
while( l <= r)
{
int mid = (l + r) / 2;
if(ans[mid] + c[i] == num)
{
node = true;
break;
}
else if(ans[mid] + c[i] < num)
{
l = mid + 1;
}
else
r = mid - 1;
}
if(node)
break;
}
if(node)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}